Artificial intelligent assistant

Monotone Class closed under complementation. Are Monotone classes always closed under complementation? I attempted to construct a counterexample however I was not able to do so. Recall the definition of a monotone class is: $C \subset P(X)$ for some set $X$ is called a monotone class if it is closed under countable union of an increasing sequence of sets. And similarly, is closed under countable intersection for a decreasing sequence of sets.

No, consider $C = \\{ c \subseteq \mathbb N \mid 0 \in c \\}$. $C \subseteq \mathcal P(\mathbb{N})$ is closed under unions and intersections (hence monotone) but not under complements, since $\mathbb N \setminus \\{0\\} \
ot \in C$.

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